We define mass of moving particle with velocity $v$ as

$\displaystyle m = \dfrac{m_0}{\sqrt{1 - \frac{v^2}{c^2} }}$

Where $m_0$ is the rest mass of the particle.

It is not possible for a particle(matter) to attain speed greater than or equal to speed of light because in that case mass of particle will not be
defined(or imaginary).

Nothing can exceed the velocity of light in accordance to Einstein's Special Theory of Relativity, except $tachyons$, according modern theoretical particle physics.

I don't know how far along you are in physics, so I will be explicit about all my assumptions.

Let's consider a photon and a free electron traveling along in space.

The initial energy of this system is given by the sum of the energy of the photon and the electron:

$E_i = E_\gamma + E_{i}$

and the final energy is given by $E_f = E_{f}$ (since the photon no longer exists after absorption)

$E_\gamma = E_{f} - E_{i}$

Conservation of momentum should hold as well, and we have $E_\gamma/c + p_{i}= p_{f}$ , or

$E_\gamma = c\left(p_{f}-p_{i}\right)$

since the momentum of a photon is given by its energy divided by the speed of light. This was shown experimentally by measuring the deflection of very thing gold sheets by incoming light beams.

If we set the two forms for $E_\gamma$ equal, we find

$p_{f}-p_{i} = \frac{E_f - E_i}{c}$

Now, the kinetic energy of a moving particle is given by $E(p) = \sqrt{{m_{0}}^2c^4 + p^2c^2}$ which is a convex function of the momentum $p$, i.e. $\displaystyle\frac{E(p_f) - E(p_i)}{c} \geq p_f - p_i$ with equality holding when $p_f=p_i$.

Therefore it is impossible to have $c\left(p_{f}-p_{i}\right) = E_{f} - E_{i}$, except when $p_i = p_f$, in other words when there is no photon absorbed.

Thank You sir for giving your time to comment. Just a thing I wanted to ask, is $E = mc^{2}$ incomplete? I guess for that reason alone you used the extended form of Einstein's Equation with relativistic momentum.

I originally started with $E=mc^2$ but found the argument less straightforward than the above. It isn't incomplete, it's just that people often use it one way or another without specifying what they're really talking about. In many places, you'll see $E=mc^2$ where the person really means $E=m_0c^2$, so they're talking about the rest energy, but they drop the zero subscript.

If we really mean $mc^2$, i.e. $m$ is not the rest mass, but is the relativistic mass, then $E=mc^2$ is the whole story. To see this, consider:

If you want to be a theoretical physicist I would start thinking about problems you'd really like to know the answers to. There are a huge number of things to potentially study in theoretical physics, from particles, to the cosmos, to the physics of living systems. But no matter problems what you end up interested in, I would suggest learning your mechanics from Kleppner and Kolenkov.

@Josh Silverman
–
Sir, I just saw the book suggested by you and found it really helpful. Currently I am a ninth grader. Shall I learn trigonometry and calculus first and then proceed to the book?
Also, Please tell me how to learn calculus at this age of 14 , or if any book can help me.
Thanks!

Show that total relativistic energy and momentum are not simultaneously conserved if such a thing happens. Since these laws cannot be violated it is not possible for a electron to completely absorb a photon.

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## Comments

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TopNewestBy conservation of momentum

$\displaystyle \dfrac{h}{\lambda } = m_e v$

$\implies$$\displaystyle \lambda = \dfrac{h}{m_e v}$ ........ (1)

Where 'v' is velocity of electron after absorbing photon and $m_e$ mass of electron

By conservation of energy

$\displaystyle \dfrac{hc}{\lambda } = \frac{1}{2} m_e v^2$ ........ (2)

Substituting value of $\lambda$ from equation (1) in (2)

$\displaystyle m_e vc = \frac{1}{2} m_e v^2$

From here we get

$\displaystyle v = 0$

Or

$\displaystyle v = 2c$

For the first one absorbing a photon and remain stationary is not possible (here in the case of "free" electron)

And clearly second one is not at all possible.

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why $v=2c$ is impossible ?

Also what is your status mean ? " 272 worth anything ? " @Krishna Sharma

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We define mass of moving particle with velocity $v$ as

$\displaystyle m = \dfrac{m_0}{\sqrt{1 - \frac{v^2}{c^2} }}$

Where $m_0$ is the rest mass of the particle.

It is not possible for a particle(matter) to attain speed greater than or equal to speed of light because in that case mass of particle will not be defined(or imaginary).

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Okay if It is not define , what happen , what occure that we can't define practically particle having speed greater than light ?

I guess whole mass will convert into Energy .. Am I right ?

Still you did not tell that what your status mean ? Is it your JEE _ Mains marks ?

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Nothing can exceed the velocity of light in accordance to Einstein's Special Theory of Relativity, except $tachyons$, according modern theoretical particle physics.

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In $\left(2\right)$, shouldn't you have $hc/\lambda + m_e(0)c^2 = \frac12 m_e(v)v^2$?

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Hi Swapnil,

I don't know how far along you are in physics, so I will be explicit about all my assumptions.

Let's consider a photon and a free electron traveling along in space.

The initial energy of this system is given by the sum of the energy of the photon and the electron:

$E_i = E_\gamma + E_{i}$

and the final energy is given by $E_f = E_{f}$ (since the photon no longer exists after absorption)

$E_\gamma = E_{f} - E_{i}$

Conservation of momentum should hold as well, and we have $E_\gamma/c + p_{i}= p_{f}$ , or

$E_\gamma = c\left(p_{f}-p_{i}\right)$

since the momentum of a photon is given by its energy divided by the speed of light. This was shown experimentally by measuring the deflection of very thing gold sheets by incoming light beams.

If we set the two forms for $E_\gamma$ equal, we find

$p_{f}-p_{i} = \frac{E_f - E_i}{c}$

Now, the kinetic energy of a moving particle is given by $E(p) = \sqrt{{m_{0}}^2c^4 + p^2c^2}$ which is a convex function of the momentum $p$, i.e. $\displaystyle\frac{E(p_f) - E(p_i)}{c} \geq p_f - p_i$ with equality holding when $p_f=p_i$.

Therefore it is impossible to have $c\left(p_{f}-p_{i}\right) = E_{f} - E_{i}$, except when $p_i = p_f$, in other words when there is no photon absorbed.

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Thank You sir for giving your time to comment. Just a thing I wanted to ask, is $E = mc^{2}$ incomplete? I guess for that reason alone you used the extended form of Einstein's Equation with relativistic momentum.

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I originally started with $E=mc^2$ but found the argument less straightforward than the above. It isn't incomplete, it's just that people often use it one way or another without specifying what they're really talking about. In many places, you'll see $E=mc^2$ where the person really means $E=m_0c^2$, so they're talking about the rest energy, but they drop the zero subscript.

If we really mean $mc^2$, i.e. $m$ is not the rest mass, but is the relativistic mass, then $E=mc^2$ is the whole story. To see this, consider:

$\begin{aligned} E &= mc^2 \\ &= \frac{m_0}{\sqrt{1-v^2/c^2}}c^2 \\ E^2 &= m_0^2\frac{c^4}{1-v^2/c^2} \\ E^2 &= m_0^2c^4 + E^2v^2/c^2 \end{aligned}$

Now, using $E=mc^2$ on the right side again, we get

$E^2 = m_0^2c^4 + m^2v^2c^2$

But $p = mv$, so finally we have

$E^2 = m_0^2c^4+p^2c^2$

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Sir, I am an aspiring Theoretical physicist. Could you please advice me some techniques which would make me even stronger in physics? Thank You!

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If you want to be a theoretical physicist I would start thinking about problems you'd really like to know the answers to. There are a huge number of things to potentially study in theoretical physics, from particles, to the cosmos, to the physics of living systems. But no matter problems what you end up interested in, I would suggest learning your mechanics from Kleppner and Kolenkov.

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Show that total relativistic energy and momentum are not simultaneously conserved if such a thing happens. Since these laws cannot be violated it is not possible for a electron to completely absorb a photon.

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@Sudeep Salgia @Nishant Rai Thank U for commenting!

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@John Muradeli

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I have no clue.

But it seems they have figured it out, in terms of energy and momentum conservations.

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@Josh Silverman Please help with your guidance!

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OK if there are no answers I like by tonight, I'll weigh in! Thanks for asking me @Swapnil Das

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